Question

It is well-known that cold air feels much colder in windy weather than what the thermometer reading indicates because of the "chilling effect" of the wind. This effect is due to the increase in the convection heat transfer coefficient with increasing air velocities. The equivalent wind chill temperature in \(^{\circ} \mathrm{F}\) is given by \([\mathrm{ASHRAE},\) Handbook of Fundamentals (Atlanta, GA, \(1993 \text { ), p. } 8.15]\) $$\begin{aligned} T_{\mathrm{equiv}}=& 91.4-\left(91.4-T_{\text {anbient }}\right) \\ & \times(0.475-0.0203 V+0.304 \sqrt{V}) \end{aligned}$$ where \(V\) is the wind velocity in \(\mathrm{mi} / \mathrm{h}\) and \(T_{\text {ambicnt }}\) is the ambient air temperature in \(^{\circ} \mathrm{F}\) in calm air, which is taken to be air with light winds at speeds up to \(4 \mathrm{mi} / \mathrm{h}\). The constant \(91.4^{\circ} \mathrm{F}\) in the given equation is the mean skin temperature of a resting person in a comfortable environment. Windy air at temperature \(T_{\text {ambient }}\) and velocity \(V\) will feel as cold as the calm air at temperature \(T_{\text {equiv. }}\) Using proper conversion factors, obtain an equivalent relation in SI units where \(V\) is the wind velocity in \(\mathrm{km} / \mathrm{h}\) and \(T_{\text {ambient }}\) is the ambient air temperature in \(^{\circ} \mathrm{C}\)

Short answer

The equivalent wind chill temperature relation in SI units is: $$ T_{\mathrm{equiv}}(^{\circ} \mathrm{C})= 33.0-\left(33.0-T_{\text {anbient }}(^{\circ} \mathrm{C})\right) \times(0.475-0.0326 V\,(\mathrm{km}/\mathrm{h})+0.304 \sqrt{V\,(\mathrm{km}/\mathrm{h})}) $$

Step-by-step solution

Write down the given equation

We are given an equation to calculate the equivalent wind chill temperature (in Fahrenheit): $$ T_{\mathrm{equiv}}= 91.4-\left(91.4-T_{\text {anbient }}\right) \times(0.475-0.0203 V+0.304 \sqrt{V}) $$ where \(T_{\text {ambient}}\) is the ambient air temperature in \(^{\circ} \mathrm{F}\) and \(V\) is the wind velocity in \(\mathrm{mi} / \mathrm{h}\).

Convert V from miles per hour to kilometers per hour

We need to find a conversion factor to change miles per hour to kilometers per hour. We know that 1 mile is equal to 1.60934 kilometers. So, the conversion factor will be: $$ \text{Conversion factor (C1)} = 1 \,\mathrm{mi} / \,\mathrm{h} \times \frac{1.60934 \,\mathrm{km}}{1\, \mathrm{mi}} = 1.60934 \,\mathrm{km} / \,\mathrm{h} $$ Thus, we can write the wind velocity \(V\) in km/h as: $$ V\,(\mathrm{km}/\mathrm{h})=V\,(\mathrm{mi}/\mathrm{h}) \times 1.60934 $$

Convert ambient temperature from °F to °C

We now need to convert the ambient temperature from Fahrenheit to Celsius using the following formula: $$ T_{\text {ambient}}(^{\circ} \mathrm{C}) = \frac{T_{\text {ambient}}(^{\circ} \mathrm{F}) -32}{1.8} $$ We should also convert the constant value 91.4°F to Celsius: $$ 91.4\, (^{\circ} \mathrm{C}) = \frac{91.4\, (^{\circ} \mathrm{F}) -32}{1.8} \approx 33.0 ^{\circ} \mathrm{C} $$

Substitute the conversions into the given equation

Now, we will substitute the conversions from Steps 2 and 3 into the given equation: $$ T_{\mathrm{equiv}}(^{\circ} \mathrm{C})= 33.0-\left(33.0-T_{\text {anbient }}(^{\circ} \mathrm{C})\right) \times(0.475-0.0203 (V\,(\mathrm{mi}/\mathrm{h})*1.60934)+0.304 \sqrt{V\,(\mathrm{mi}/\mathrm{h})*1.60934}) $$

Simplify and express the equivalent relation in SI units

Finally, we will simplify the equation and express it in SI units: $$ T_{\mathrm{equiv}}(^{\circ} \mathrm{C})= 33.0-\left(33.0-T_{\text {anbient }}(^{\circ} \mathrm{C})\right) \times(0.475-0.0326 V\,(\mathrm{km}/\mathrm{h})+0.304 \sqrt{V\,(\mathrm{km}/\mathrm{h})}) $$ Now, the equivalent relation is expressed in SI units, with wind velocity in km/h and ambient air temperature in °C.